Acta Mathematica et Informatica Universitatis Ostraviensis

Kazimierz SzymiczekKnebusch-Milnor exact sequence and parity of class numbers

Acta Mathematica et Informatica Universitatis Ostraviensis, Vol. 4 (1996), No. 1, 83--95

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Acta Mathematica et Informatica Universitatis Ostraviensis 4(1996)83-96 83

Knebusch-Milnor Exact Sequence and Parity of Class Numbers

KAZIMIERZ SZYMICZEK

A b s t r a c t . We show that the Knebusch-Milnor exact sequence for the Witt group of quadratic forms over a number field K can be used to determine the parity of ideal class number of the field K. As an example we reprove the classic results on quadratic number fields with odd class numbers.

1991 M a t h e m a t i c s S u b j e c t Class i f icat ion: Primary 11R29, Secondary 11E12

Introduction

The relat ionship between the integral theory of quadra t ic forms and class numbers of algebraic number fields is one of the fundamental features of algebraic number theory. It manifests most clearly in the Gauss ' determinat ion of the 2-rank of the (narrow) ideal class group of a quadra t i c field. On the other hand , the simpler theory of quadra t i c forms over number fields seems to have looser connection to the subtle quest ions of class numbers and class group s t ructure .

In this note , however, we show tha t a general result on W i t t groups of quadra t ic forms over number fields, the exactness of the Knebusch-Milnor sequence, supplies an effective criterion for the par i ty of class numbers . We try the s t rength of the criterion by showing t h a t it provides a complete determinat ion of quadra t ic number fields with odd class numbers . Thus these subtle ar i thmet ic results follow ra ther easily from the simpler rational (as opposed to the more sophist icated integral)

quadra t ic form theory.

The fact t h a t W i t t rings of number fields carry some d a t a on the par i ty of class numbers was first noticed in [7], and then some further results in this direction were given in [4]. A deeper s tudy of the relations between W i t t equivalence of number fields and 2-ranks of ideal class groups will be presented in a forthcoming paper by P . E. Conner, R. Perlis and the au thor [2].

The no ta t ion is s t andard . We write ( a i , . . . , an) for the n—dimensional diagonal quadra t ic form arXl + • • • + anXl over a field K, and DK{ax,..., an) for its value set. The Legendre symbol is wr i t ten {a/p).

84 K.Szymiczek

1 The Knebusch-Milnor sequence

Let K be a number field, OK its ring of integers and C(K) the ideal class group of K. We write Q(I\) for the set of all finite primes of K. For each p £ Q(K), let rip be the p —adic completion of Ii, and Tip the residue class field of Zip. Then, with p running over all finite primes of K we have the following Knebusch-Milnor sequence for the Witt groups W(OK), W(K) and W(Kp ) :

0 -> W(OK) A VV(K) -^> JJ W(K^) --> C(K)/C(K)2 -> 0. (1) P

Here i is the natural injection. In fact, without going into details of Witt groups of rings, it is best for us to define W(OK) as the kernel of the homomorphism dK.

We recall now the definition of 8K • For each finite prime p of K we consider the second residue class homomorphism

dp : W(Kp) —+• W(Kp~).

This can be defined only after fixing a prime element IT in i i p . Then every element </> £ W(Kp) can be written as

<j> = (au . ..,afc,bi7r, . . .,6m7r),

where a2-,b?- are units in Kp. We set

dp(<j>):=(bu...,bm)€W(Kp~),

where b is the canonical image of the p—adic unit b in the residue class field Kp. Notice that this construction does not distinguish between dyadic and nondyadic primes.

Now we globalize. Given <f> £ W(K) we write </>p for the image of </> under the natural homomorphism W(K) —> VV(Arp) and define <9p(</3) := <9p(^p). Then, for each prime p we have the group homomorphism

dp : W(K) -> VV(Kp).

For any fixed (j) £ W(K) we have dp(<f>) = 0 for almost all primes p . Hence we can aggregate these homomorphisms into one homomorphism

dK:W(K)-^]JW(Ep)) 0*M = (dpM). P

dK is said to be the boundary homomorphism. We will write d inste.cid of dK)

when there is no risk of ambiguity

It remains to recall the definition of A. Let rj = (rfp) £ [ J p W(Kp ). We set

A(^) = [ n p e ( " P ) ] C ( A ' ) 2 . p

Knebusch-Milnor Exact Sequence and Parity of ("lass Numbers 85

Here e : W(Kp ) —» TLjTL is the dimension index homomorphism, and the square brackets are used to denote the ideal class in C(K),

The main result we will use in this paper states that (1) is an exact sequence. In fact, the sequence is exact independently of the particular choices of the in­volved second residue class homomorphisms. Proofs can be found in the books by J. Milnor and D. Husemoller [5], and W. Scharlau [6].

2 Knebusch-Milnor sequence class number parity criterion

The following criterion for parity of the class number of K follows immediately from the exactness of the sequence (1).

Class n u m b e r parity criterion. The class number J%K of a number field K is odd if and only if the boundary homomorphism 8K is surjeetive.

P R O O F : The class number IIK is odd if and only if the ideal class group C(K) satisfies C(K) = C(K)2 if and only if A is a trivial homomorphism if and only if OK is surjeetive. •

The subjectivity of 8 — 8K will be established when we show that a set of genera­tors of the target group lies in the image of 8.

The group J_Jp !V(I\p ) is the direct sum of cyclic groups of orders 2 and 4, and we will choose the generators for these cyclic direct summands as our set of generators for the group. To explain this we first recall the structure of the Witt groups of residue class fields Kp. Here I\p is a finite field and the structure of the Witt group VV(I\p ) depends on the number of element^ N(p) :-= | N p | of the residue class field (rather than on the characteristic).

If N(p) is even (so that char I\p = 2), the Witt group W(Kp ) is cyclic of order 2 generated by (1). If N(p) = 3 (mod 4) the Witt group W(Kp ) is cyclic of order 4 generated by (1). And for N(p) = 1 (mod 4) the Witt group W(Kp ) is the Klein four-group generated by (1) and (u), where u is the residue (mod p) of a nonsquare unit u in I\p. In the case N(p) = 1 (mod 4) we choose and fix a nonsquare unit u G Kp and denote it u = u(p).

From this it follows that the group J J p W(Kp ) is the direct sum of cyclic groups generated by the elements in the set

{77P :pef l( jR.)} U {p11^ : p e Q(K) with N(p) = 1 (mod 4)},

where

{ГÌ )* - 1 (0) for q ^ p ,

86 K.Szymiczek

(uUW)a ~ í ^ f°Г q = P ' ^ )Ч ~ { (0) for q ^ p .

Here the subscript q indicates the q—th coordinate of the element.

The following lemma shows that trying to establish the surjectivity of d we do not have to bother about locating the generators Liu(P) in the image of d.

L e m m a 1. For a prime p G ft(K) with N(p) = 1 (mod 4),

77P G im d iff //W(P) G im d.

Hence the boundary homomorphism d is surjective if and only if all the generators 77P, p £ -3(-K), lie in the image of d.

PROOF: This is a simple consequence of the exactness of the sequence (1). For observe that the generators 77P and L*u(P) have the same image under A :

\(vP) = \p]C(Kf = \(^).

Hence, by the exactness of the sequence (1),

77P G im (9 iff 77P G ker A * iff pu^ G ker A iff /iu(P) G im d

which proves the Lemma. •

L e m m a 2. Let p be a rational prime and suppose p is inert in K, that is, POK — P is a prime ideal. Then for the one-dimensional form (p) G W(K) we have

d(P) = vp

for a suitable choice of the residue class homomorphism dp. Hence 77P G im d for all inert primes p of K.

P R O O F : Choosing p as the prime element in the completion I\p it is immediate that d(p) =rjP. D

L e m m a 3. For an arbitrary prime ideal p in OK the following are equivalent. (a) 77P G imd.

(b) There is a number (3 G OK and an ideal a in OK such that

(30K=P*2. (2)

(c) There is a number (3 G OK such that with an appropriate choice of the prime element in the p—adic completion of K, we have

rjP = d((3)

Knebusch-Milnor Exact Sequence and Parity of Class Numbers 87

for the one-dimensional form (/?) in W(K).

P ROO F: If 77P £ im<9, then, by the exactness of the sequence (1), we have rjP £ ker A, that is,

C{Kf = A(-/P) = [V)C{K)\

Thus there are a number (5 E OK and an ideal a of OK such that (2) holds. This proves (a) => (b).

Now assume (b) holds and consider the 1-dimensional form ((3) E W(K). By (2), we have dq(/3) — 0 for all prime ideals q ^ p . On the other hand, (3 has odd p—order, hence f3 ~ TTX2 for some 7r, X E Kp with ordp IT — I. Choosing IT as the prime element in Kp and adjusting the residue class homomorphism dp to the new choice of the prime element, we get dp(/3) = (1). This shows that the new boundary homomorphism d (modified at the p—th coordinate) satisfies d(/3) = 77P. This proves (b) => (c), and (c) =-> (a) is trivial.

3 Quadratic fields with odd class numbers

The following is the complete list of quadratic fields with odd class numbers:

K = Q(v/rf), d = - l , ± 2 , - g , r , 2 < l , a g i , (3)

where r is an arbitrary odd prime and q and q\ are primes congruent to 3 (mod 4) (see [1], Corollary (18.4)).

We now give a new proof of this result based on the class number parity criterion discussed in Section 2, and on the techniques of rational quadratic form theory.

Theorem. For a quadratic number field K the following statements are equiva­lent.

(A) The class number hK is odd.

(B) The boundary homomorphism 8K is surjective.

(C) Every rational prime number p which is split or ramified in K is represented, up to the sign, as the norm of a number in K.

(D) K = Q ( v J ) , where d = — 1, ±2, — q, r, 2a, qqi, with r an arbitrary odd prime number and q and q\ prime numbers congruent to 3 (mod 4).

The proof will be given in a series of propositions showing that (A) iff (B )=»(C)=>(D)=»- (C)=>(B) .

K.Szymiczek

We will use the quadratic form theory over the p—adic fields and the local-global principle for quadratic forms over the rational number field Q. Another often used argument is the theorem on the existence of prime numbers p with prescribed values of Legendre symbols ( a i / p ) , . . . , (a^/p) for any set a i , . . . , ak of integers multiplicatively independent in the square class group Q*/Q*2 (see [3], Satz 147).

P ropos i t ion 1. (A) iff (B).

P R O O F : This is our class number parity criterion from Section 2. D

Propos i t ion 2. (B) => (C).

P R O O F : Suppose p is a rational prime split or ramified in K. Let POK = PP, where p is a prime ideal in OK (here p = p when p ramifies in K). Since OK is surjective, rp £ im<9. Hence, by Lemma 3, there is a /3 £ OK and an ideal a in OK satisfying

/30K = pa 2 and pOK = pa 2 .

It follows that, for a positive rational integer a,

^ O K = p p - ( a a ) 2 = p a 2 0 K .

Hence NK/q(f3/a)OK = POK, and p = ±NK/®(P/O>), as required. D

P ropos i t i on 3 . (C) => (D).

PROOF: Let K = Q(Va) , where d ?- 1 is asquarefree integer and let K satisfy (C). We write (1, —d) for the norm form X2 — dY2 in K/Q. Claim 1. If d is divisible

by at least two distinct primes, then all odd prime factors of d are congruent to 3 (mod 4). Suppose d is divisible by at least two distinct primes q and r and assume q = 1 (mod 4). Choose a prime p satisfying

(q/p) = (r/p) = - 1 and (c/p) = 1,

where c satisfies d = qrc. Then we have (d/p) = 1 and so p splits in K. We will show that neither p nor — p is a norm from K.

By reciprocity, (p/g) = (q/p) = — 1. Hence p is a nonsquare unit in Qq. Then it follows that p is not represented by the form (1,— d) over Qq, hence p is not represented by the norm form (1, — d) over Q.

On the other hand, —1 is a square in Qqi hence p and —p are in the same square class in Qq. Since p is not represented by the form (l,—d) over Qq, also —p is not represented. Hence neither p nor —p is a norm from A', contrary to the hypothesis (C). This proves Claim 1.

Claim 2. If d > 0, then d is divisible by at most two distinct primes. For suppose

Knebusch-Milnor Exact Sequence and Parity of Class Numbers 89

Q^Qi^Q2 are distinct prime factors of d and assume that q and q\ are odd primes. Choose a prime p such that

(-1/D) = (q/p) = (q2/p) - - 1 and (r/p) = 1,

for all prime factors r of d distinct from q and q2. Hence we have p = q = q\ = 3 (mod 4), the latter by Claim 1. Moreover, (d/p) = 1 so that p splits in K. By quadratic reciprocity,

i-p/q) = - (p /g) = (flf/p) = - i ,

hence — p is a nonsquare unit in Qg and so is not represented by the form (1, — d) over Qg . On the other hand, (qi/p) — 1, hence (p/qi) = — 1 and so p is not represented by (1, — d) over Q g i .

It follows that neither — p nor p is represented by the norm form (1, —J) over Q, contradicting (C). This proves Claim 2.

Claim 3. If d < 0, then either d = —1,-2, or — d is a prime congruent to 3 (mod 4). Suppose —d > 2 and d has two distinct prime factors q and q\. Suppose

that q is odd. Write d = — qq\c, where c is a positive integer. We choose an odd prime p satisfying

(Q/P) = (qi/p) = - 1 , (~1/P) = 1 and (c/p) = l.

Then (d/p) = 1 so that p splits in K. Moreover, p = 1 (mod 4) and (g/D) = — 1 imply that (p/o) = (q/p) — — 1. Thus p is a nonsquare unit in Qg and so is not represented by (1, — d) in Qg . It follows that p is not represented by the norm-form (1,—J) over Q. Since d < 0, the form (1,— d) represents only positive rational numbers, hence — p is not represented either, contradicting (C).

Thus we have proved that —d = q is a prime number. It remains to show that q ™ 3 (mod 4). Suppose q ~ 1 (mod 4) and choose a prime number p satisfying

( - - / » = (?/p) = - I -

Then (d/p) = 1 and (p/q) = (<//p) = — 1. It follows that p splits in K and is not represented by the form (1,—J) over Qq. Hence p is not the norm from K, and -p is not either. This again contradicts (C) and proves our claim. Summarizing,

if d > 0, then by Claim 2, d is of the form r,2q or goi, where r,q,q\ are prime numbers. By Claim 1, q and q\ are congruent to 3 (mod 4). On the other hand, when d < 0, then Claim 3 gives the desired result. D

Proposition 4. (D) => (C).

Our argument splits into two cases depending on the number of ramified primes in the field K. Hence it will be convenient to separate the fields with one ramified

90 K. Szymiczek

prime from those with two ramified primes. We defme the following two classes of quadratic fields:

A := {Q(V5), c ř = - l , ± 2 , - g , r , g EE 3 (mod 4), r = l (mod 4)},

B := {Q(VS), d=q,2q,qqu q = qt = 3 (mod 4)},

where always g, gi and r are rational prime numbers.

Thus AUB covers all fields in the list (3), and in each field of the class A there is just one ramified prime, while in fields of the class B there are two ramified primes.

The following two lemmas provide a more detailed version of Proposition 4.

L e m m a 4. Let K — Q(vd) be a quadratic number field listed in (3). (a) //

K belongs to the class A, then every rational prime split in K is the norm of a number in K. (b) If K belongs to the class B and p is a rational prime split in K,

then either p or —p is the norm of a number in K.

P R O OF: We want to show that for every split rational prime p the norm form (1, — d) represents one of dbp over Q. For this we use the local-global principle. It suffices to show that

pEDqp{l,-dy for P\2pd (4a)

in the čase (a), and

p£Dqp{l)-d) for P\2pd or - p £ D Q P ( 1 , -d) for P\2pd (4b)

in the čase (b). Here P denotes a rational prime. Consider hrst the prime P = p.

If p ^ 2 is split in Ji, then we háve (d/p) = 1, and, by HensePs lemma, d is a square in Qp. If p — 2 splits in /i, we must háve d = 1 (mod 8). But then d is a square in the dyadic field Q2. Hence if any prime p splits in K, then oř is a square in Qp. It follows that (1, — d) ~ (1,-1) over Q p , and so the form (1, —d) is universal over Qp, hence represents p as well as — p over Qp. Thus when P — p the statements (4a) and (4b) hold in the stronger form asserting that both p and —p are represented by (1, —oř) over Q p . Now we proceed to the statement (a). The

prime P = p has already been dealt with and so to prove (4a) it suffices to show that for any split prime p,

P G % ( 1 , - ( I ) for P\2d. (5)

Consider the prime P = q = 3 (mod 4). First assume that p ^ 2. If q \ d, we must háve oř = — g, and since p is split in K, the field discriminant — g is a quadratic residue modp. Hence, in terms of Legendre symbols, (q/p) = (— l/p) = ( — 1 ) ( P - 1 ) / 2 . On the other hand, by quadratic reciprocity,

(p/q) = (q/p) • ( - 1 ) ^ ' v = (q/p) • ( - 1 ) ^ = 1.

Knebusch-Milnor Exact Sequence and Parity of Class Numbers 91

Hence each split rational prime p / 2 is a quadratic residue modg. By Hensel's lemma, p is a square in Qq and so p is certainly represented by the form (1, —d)

Now let p = 2. So 2 is split in K = Q(Va) , with J = —g. This forces —g s 1 (mod 8), hence (2/q) = 1, and so 2 is a square in Qq. Thus p = 2 is represented by the form (V—d) over Qg. This establishes (5) in the case P = q. Next we consider P = r. Now d = r = 1 (mod 4). Again we first assume that p ^ 2. Then (r/p) = 1, and by reciprocity, (p/r) = 1. Hence p is a square in Q r and so p is represented by the form (1, —d) over Q r .

When p = 2 is the split prime, we have d = r == 1 (mod 8), and it follows that 2 is a square in Q r . Then p = 2 is represented by the form (1, —J) over Q r . Thus

for all d considered in case (a) we have verified (5) for all primes P \ 2d except for P = 2. In other words, the Hilbert symbol (p,J)p assumes the value 1 for all primes P (including the infinite primes) except possibly for P = 2. By Hilbert reciprocity, (p, d)2 = 1, as well. It follows that p is represented by the form (1, —d) everywhere locally, hence, by the Hasse-Minkowski Principle, p is represented by (1, —J) over Q, as desired. Now we prove (b).

Let p be a rational prime split in K. Since (4b) has already been verified for the prime P = p with or replaced by and, it suffices to show that

PeDqp(lJ-d) for F|2d or - p G A Q P ( 1 , -q) for P \ 2d. (6)

Consider the prime P = q. We have

(-p/q) = (-i)^(p/q) = -(p/q),

hence either p or — p is a square in Qq. It follows that one of the numbers dtp is represented by the form (1, —J) in the field Qq.

Thus in the cases d = q and d = 2q we have proved that one of the numbers ±p is represented by the form (1, —d) in all the fields Qp, except possibly in Q2.

Now we consider the case d = qqi. The same argument as for P = a above applies when P = oi. We show that one of dbp is simultaneously represented by (1,—J) over Qg and over Qqi. We again consider separately the cases p ^ 2 and p = 2.

If p 9-: 2, then p split in K implies that (qqi/p) = 1, hence (q/p) = (gi/p). On the other hand,

(p/q) = (q/p) • ( - 1 ) ^ ' ^ = (qi/p) • ( - l ) ^ 1 1 ^ = (p/ ? 1 ) ,

and also (—p/q) = —(p/q) = —(p/qi) = (—p/qi)- Hence either p is a square in Qg

and in Q?1 or —p is a square in Qq and in Q g i . It follows that when p ^ 2, one of the numbers ± p is represented by the form (1, — d) in Qq and in Q(?1. Now let p = 2 be the split prime. Then we have d = qqi = 1 (mod 8), and so g = gi (mod 8).

92 K.Szymiczek

If q = q\ = 3 (mod 8), then (-2/q) = (~2/q\) = 1, hence - 2 is a square in both Q . a n d Q ^ .

The other possibility is q = gi = 7 (mod 8), and then (2/g) = (2/q\) = 1, hence 2 is a square in both Qq and Q g i .

In each case either p = 2 or — p = — 2 is represented by the form (1, —d) over Qg

and over Qqi. Summarizing, we have shown that (6) holds for all prime divisors

P of 2d except possibly for P = 2. By Hilbert reciprocity, (6) holds for P = 2, as well.

Thus with a proper choice of the sign, ±p is represented by the form (1,—d) everywhere locally, hence, by the Hasse-Minkowski Principle, ±p is represented by (1, —J) over Q, as desired. •

L e m m a 5. Let K = Q(vd) be a quadratic number field listed in (3). Let p be a rational prime ramified in K. Then either p or —p is the norm of a number in K.

P R O O F : When K E A, then we have NK/Q(V71) = —d for d = ±2, —g, r and NK/Q(1 -f i) = 2 for d = —1. This proves the Lemma for fields in A. Now we assume that K G B. Consider first the case when d = q. Then the ramified primes are p = 2 and p = a. If a = 3 (mod 8), then (—2/q) = 1, hence, by the local-global principle, —2 is represented by the form (1, — q) over Q. If q = 7 (mod 8), then (2/q) = 1, hence 2 is represented by the form (1, —g) over Q. Thus either 2 or —2 is the norm of a number in K. As to the ramified prime p = g, we have —g = NK/Q(-^)- NOW let J = 2g. The ramified primes are p = 2 and p = g. If

g = 3 (mod 8), then (—2/g) = 1, hence —2 is a square in Qq. It follows that the quadratic form (1, — 2g) represents both —2 and g over Qq) and by the local-global principle, the form represents —2 and q over Q.

If q = 7 (mod 8), then (2/g) = 1, hence 2 is a square in Qq. It follows that the quadratic form (1, —2g) represents both 2 and —g over Qq, and by the local-global principle, the form represents 2 and — q over Q. Finally let d = ggi so that the

ramified primes are p = g and p = gi. First suppose that (g/gi) = V Then g is a square in Q g i , hence is represented by the form (1,—ggi) over Qqi. Moreover, (—gi/g) = — (gi/g) = (q/q\) — 1, hence — q\ is a square in Qq. It follows that (1, —ggi) = (1, g) over Qg , and so g is represented by the form (1, —ggi) over Qq

as well. By the local-global principle, q is represented by the norm form (1, —ggi) over Q.

Now consider the other ramified prime p = gi. We know that g is a square in Qqi, hence (1,— ggi) = (1,— gi) over Qqi so that —gi is represented by (1,— ggi) over Qqi. On the other hand, —gi is a square in Qq and so —gi is represented by the form (1,—ggi) over Qq. By the local-global principle, —gi is represented by the norm form (1, -qqi) over Q.

Thus if (g/gi) = 1, the numbers g and - g i are represented by the norm form of

Knebusch-Milnor Exact Sequence and Parity of Class Numbers 93

K, hence are norms of some numbers in K. It remains to consider the case when

(q/qi) — — 1. But then we have (q\/q) = —(q/q\) = 1, and by symmetry, — q and oi are the norms of some numbers in K. •

P R O O F OF PROPOSITION 4: Combine Lemmas 4 and 5. •

In the final part of the proof we will use the following auxiliary result.

Lemma 6. Let K be a quadratic number field. Suppose p is a rational prime and

±pa2 = M, (7)

where a E N, j3 E OK and the rational positive integer a assumes the smallest possible value.

(a) Ifp is split in K, then (3 and j3 do not have any common prime ideal factors.

(b) If p is ramified in K, POK = q2, where q is a prime ideal, then the only common prime ideal factor of /3 and J3 is q and

ordq/? = ordq/3 = 1.

P R O O F : Let p be a rational prime which is either split or ramified in K, and let p be a prime ideal in OK and

p|/? and p|/i, (8)

where (3 and /? satisfy (7). There are three cases to consider. First assume that the prime ideal p is a factor of a split rational prime r. Since p|/J implies p|/?, we have rOK = PP|/2- It follows that r|/5 and so also r\/3. This forces r|a and so, with Pi — f3/r and a\ — a/r, we have

±pa? = /?i&,

where a\ < a and j3\ E OK, contrary to the choice of a and /?. The second case is

when p is an inert prime. Hence p = r(9x, where r is a rational prime. Then (8) implies that r\/3 and r\/3, hence r|a, and we argue as in the first case to reach a contradiction. Thus far we have proved that in both cases (a) and (b) the numbers

/3 and (} do not have any split or inert common prime ideal factors.

Now we consider the case when p is the factor of a ramified prime a, that is, p 2 = qOK- Then (8) implies that q\pa2. Here we have to distinguish between the

cases (a) and (b).

If p is split, then we must have q | a2 (since q is ramified and so p # q)-

94 K.Szymiczek

If p is ramified in K, POK = q2, where q is a prime ideal, and p / q, then q\pa2 implies again that g|a2. In both cases it follows that g|a, hence p4\P/3. Since

p = p, it follows that p2\P and p2 |/5, hence q\a) q\/3, q\/3, and once again we reach a contradiction. This finishes the proof of (a) and shows that, in the case (b), the

only common prime ideal factor of /? and J3 is the ideal q. Suppose q2\/3. Since pOK = q2 and q = q, we also have q2|/3. Hence p\f}y p\/3 and p2\PP = ±pa2. It follows that p|a, and, as above, this contradicts the choice of a and j3. Hence ordq j3 = 1, and since q = q, we have also ordq /? = 1.

P ropos i t ion 5. (C) => (B).

PROOF: In view of Lemmas 1 and 2 it suffices to show that for all split and all ramified primes p of the field K, the generators r/P are in the image of d. So let

p be a factor of a rational prime p which splits or ramifies in K. By (C), we can assume that there is a positive rational integer a and a number p G OK such that

dbpa2 = pp.

Of all equations of this type we choose one in which a assumes the least value.

We know from Lemma 6 how to determine the common prime ideal factors of p and p. Hence we can compute easily the images of the quadratic form (P) G W(K) under all residue class homomorphisms. First assume that p splits in K. Let

POK — PP be the prime ideal decomposition of the prime p and let the notation be chosen so that p | p. Then, by Lemma 6, we have

ordp P = ordp pp = ordp pa2 = 1 (mod 2).

Thus P = nx2, where IT is a prime element in Kp and x G Kp. We choose 7T to be the prime element determining the action of the second residue class homomorphism dp. Then we have dp(P) = (1).

On the other hand, ordp /? = 0, hence dp(p) = 0. Now we compute 3q(/?), when

q is any prime ideal in OK distinct from the prime factors p and p of p. Suppose q\p. Then, by Lemma 6, we have

ordq P = ordq P/3 = ordq pa2 = ordq a2 = 0 (mod 2).

Clearly, if q does not divide /?, then we also have ordq/? = 0 (mod 2). Thus we have dq(P) = 0 for all q ^ p . Hence d(p) = (<9q(/?)) = rp. It remains to consider

the ramified primes. Let POK = q2 be the prime ideal decomposition of the prime p. Then, by Lemma 6, we have

ordq P — ordq P = 1.

Knebusch-Milnor Exact Sequence and Parity of Class Numbers 95

Thus /? is a prime element in I\q and so it determines the action of the second

residue class homomorphism <9q. With this choice of the prime element we have

dq(j3) = (I). Now we compute <9p(/i), when p is any prime ideal in OK distinct

from the prime factor q of p. Suppose p|/3. Then, by Lemma 6, we have

ordp P = ordp /3/3 = ordp pa2 = ordp a2 = 0 (mod 2).

Clearly, if p does not divide /?, then we also have ordp (3 ~ 0 (mod 2). Thus we

have 9p(P) = 0 for all p ?- q. Hence d(/3) = (dp(/3)) = r]^.

This proves that the boundary homomorphism d is surjective. D

References

[1] Conner, P. E., Hurrelbrink, J., Class number parity, World Scientific Publishing Co., Singapore, 1988.

[2] Conner, P. E., Perlis, R., Szymiczek, K., Wild sets and 2-ranks of class groups, Acta Arith., to appear.

[3] Hecke, E., Vorlesungen über die Theorie der algebraischen Zahlen, Akademische Ver-lagsgesellschaft, Leipzig, 1923.

[4] Jakubec, S., Marko, F., Szymiczek, K., Parity of class numbers and Witt equiva-lence of quartic ftelds, Math. Comput. 64 (1995), 1711-11715. Corrigendum, ibid. (to appear).

[5] Milnor, J., Husemoller, D., Symmetric Bilinear Forms, Springer-Verlag, Berlin-Heidelberg-New York, 1973.

[6] Schaгlau, W., Quadratic and Hermitian Forms, Spгinger-Verlag, Berlin-Heidelberg-New York, 1985.

[7] Szymiczek, K., Witt equivalence of global fields, Commun. Alg. 19 (1991), 1125-1149.

Address: Instytut Matematyki, Uniwersytet Šlaski, Bankowa 14, 40 007 Katowice, Poland, e-mail: szymiczekgate.math.us.edu.pl

Talk at the First Czech-Polish Conference on Number Theory, University of Ostrava, Ostrava, Czech Republic, May 30, 1996.